Analysis Problem of the Day 76

Today’s problem appeared as Problem 9 on the UCLA Fall 2023 Analysis Qual:

Problem 9. a) Let $f$ be an entire function, $c_1,c_2, \omega_1,\omega_2 \in \mathbb{C}$ such that $\omega_1,\omega_2$ are linearly independent over $\mathbb{R}$ and

$f(z+\omega_1)=f(z)+c_1, \quad f(z+\omega_2) = f(z)+c_2,$

for all $z \in \mathbb{C}.$ Show that $f(z)$ must be a linear function, i.e. $f(z)=az+b.$

b) For $a,b \in \mathbb{R},$ let

$R_a:=\{z \in \mathbb{C}: 0 \leq \text{Re}(z) \leq a; 0 \leq \text{Im}(z) \leq 1\}$

be an $a \times 1$ rectangle with the left corner at the origin, and suppose that $\phi:R_a \to R_b$ is a homeomorphism that preserves the sides of the rectangles which is also holomorphic on the interior of $R_a.$ Show that $a=b.$

Solution: a) Since $\omega_1,\omega_2$ are linearly independent over $\mathbb{R},$ every complex number $c$ can be written as $c= a_1 \omega_1 + a_2 \omega_2, a_1, a_2 \in \mathbb{R}.$ Now, taking derivatives on both sides yields $f'(z+\omega_1)=f'(z+\omega_2)=f'(z)$ for all $z \in \mathbb{C}.$ Thus, if $M = \sup_{P} |f'(z)|,$ where $P$ is the closed parallelogram cell with vertices at $0, \omega_1, \omega_2, \omega_1 + \omega_2,$ respectively, then

$|f'(c)| = |f'(\{a_1\}\omega_1+\{a_2\}\omega_2)| \leq M,$

where $0 \leq \{a\} <1$ is the fractional part of $a \in \mathbb{R}.$ Thus, $|f'(z)| \leq M,$ i.e. $f'$ is a bounded entire function, so that $f'(z)=c$ is constant. It follows from the Fundamental Theorem of Calculus that $f(z)=az+b$ is a linear function.

b) A very intuitive idea based on the claim in (a) would be to extend $\phi$ to a function with some periodicity in both directions and then use the conclusion in (a). However, the issue with that approach is that gluing the function $\phi$ might not be possible along the sides, since the homeomorphisms of, say, the top and bottom sides need not be the same. However, we may rectify this situation with a reflection trick, similar to the idea of the Schwarz reflection principle. Indeed, let $f_1:R_{2a} \setminus R_a \to R_{2a}, f_2: R_{2b} \setminus R_b \to R_{2b}$ be the maps $(x,y) \to (2a-x,y), (x,y) \to (2b-x,y)$ (which one can verify are antiholomorphic). Then, $f_2 \circ \phi \circ f_1$ is a holomorphic function on the interior and glues continuously to $\phi$ on the line segment $a+it, 0<t<1.$ Thus, applying Morera’s theorem as in the proof of the Schwarz reflection principle yields an extension $\phi_1:R_{2a} \to R_{2b}$ that is a side-preserving homeomorphism and which is holomorphic on the interior. Repeating this procedure horizontally thus extends $\phi$ to a map $\Phi$ holomorphic on the strip $0 < y < 1,$ and extending similarly in the vertical direction yields an entire map $\Psi$ extending $\phi.$

We now verify that $\Psi$ satisfies the hypotheses in (a). Indeed, on $4R_a,$ we have that if $R_1,R_2,S_1,S_2,$ are the reflections across $x=2a,x=4a,y=2b,y=4b,$ respectively, then $S_2 \circ S_1 \circ \phi \circ R_1 \circ R_2 = \phi$ by construction. In particular, one may check that $R_1 \circ R_2 = z-2a, S_2 \circ S_1 = z+2b.$ It follows that $\phi(z-2a)=\phi(z)+2b,$ and similarly, $\phi(z-2i)=\phi(z)+2i.$ Thus, $\phi$ satisfies the hypotheses in (a), so $\phi(z)=a_1z+a_2,$ which immediately implies $a_2=0$ and $a_1 = 1,$ so that $a=b$ and $\phi(z)=z$ is the identity.

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